Question: Solve for $z$, $ -\dfrac{4z + 3}{3z + 2} = \dfrac{3}{3z + 2} - \dfrac{4}{12z + 8} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z + 2$ $3z + 2$ and $12z + 8$ The common denominator is $12z + 8$ To get $12z + 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{4z + 3}{3z + 2} \times \dfrac{4}{4} = -\dfrac{16z + 12}{12z + 8} $ To get $12z + 8$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{3}{3z + 2} \times \dfrac{4}{4} = \dfrac{12}{12z + 8} $ The denominator of the third term is already $12z + 8$ , so we don't need to change it. This give us: $ -\dfrac{16z + 12}{12z + 8} = \dfrac{12}{12z + 8} - \dfrac{4}{12z + 8} $ If we multiply both sides of the equation by $12z + 8$ , we get: $ -16z - 12 = 12 - 4$ $ -16z - 12 = 8$ $ -16z = 20 $ $ z = -\dfrac{5}{4}$